\newproblem{lay:1_3_3}{
   % Problem identification
	 \begin{large}
 	   \hspace{\fill}\newline
     \textbf{Lay, 1.3.3}
	 \end{large}
	 \\
   \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

   % Problem statement
   Draw in a graph $\mathbf{u}$, $\mathbf{v}$, $-\mathbf{v}$, $-2\mathbf{v}$, $\mathbf{u}+\mathbf{v}$, $\mathbf{u}-\mathbf{v}$, and $\mathbf{u}-2\mathbf{v}$ with 
	 $\mathbf{u}=\begin{pmatrix}-1 \\ 2\end{pmatrix}$ and $\mathbf{v}=\begin{pmatrix}-3 \\ -1\end{pmatrix}$
}{
   % Solution
   Let's make first all these calculations:\\
	 $\begin{array}{rcl}
	   \mathbf{u}&=&\begin{pmatrix}-1 &  2\end{pmatrix}\\
		 \mathbf{v}&=&\begin{pmatrix}-3 & -1\end{pmatrix}\\
		 -\mathbf{v}&=&-\begin{pmatrix}-3 & -1\end{pmatrix}=\begin{pmatrix}3 & 1\end{pmatrix}\\
		 -2\mathbf{v}&=&-2\begin{pmatrix}-3 & -1\end{pmatrix}=\begin{pmatrix}6 & 2\end{pmatrix}\\
		 \mathbf{u}+\mathbf{v}&=&\begin{pmatrix}-1 & 2\end{pmatrix}+\begin{pmatrix}-3 & -1\end{pmatrix}=\begin{pmatrix}-4 & 1\end{pmatrix}\\
		 \mathbf{u}-\mathbf{v}&=&\begin{pmatrix}-1 & 2\end{pmatrix}-\begin{pmatrix}-3 & -1\end{pmatrix}=\begin{pmatrix}2 & 3\end{pmatrix}\\
		 \mathbf{u}-2\mathbf{v}&=&\begin{pmatrix}-1 & 2\end{pmatrix}-2\begin{pmatrix}-3 & -1\end{pmatrix}=
		   \begin{pmatrix}-1 & 2\end{pmatrix}+\begin{pmatrix}6& 2\end{pmatrix}=\begin{pmatrix}5 & 4\end{pmatrix}\\
	 \end{array}$\\
	 The following figure shows these vectors
	\begin{center}
		\includegraphics[width=10cm]{Tema1/lay_1_3_3.eps}
	\end{center}
}
\useproblem{lay:1_3_3}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
